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Home›Probability and statistics›Probability

NOT, OR and AND

The chance of at least one six in four rolls of a die is 1 − (5/6)⁴ = 671/1296, about 0.518.

Two dice bets that looked the same in 1654

BackgroundIn the summer of 1654 Blaise Pascal wrote to Pierre de Fermat about questions put to him by Antoine Gombaud, the Chevalier de Méré. One concerned two bets. Betting on at least one six in four throws of a die was a winning bet; Pascal wrote that the advantage is "as 671 to 625". Betting on at least one double six in 24 throws of two dice seemed like the same bet scaled up, because 24 is to 36 as 4 is to 6. Yet it lost. The complement rule explains it: 1 − (35/36)²⁴ is about 0.491, just under one half. The letters between Pascal and Fermat that summer, mostly about how to split the stakes of an unfinished game, are usually counted as the start of probability theory.

Counting outcomes one by one works for one die or two. De Méré's first bet already has 6 × 6 × 6 × 6 = 1,296 outcomes, and the second has far more. Three rules handle most questions of this size: the complement rule for NOT, the addition rule for OR, and the multiplication rule for AND. They answer small questions too, such as the chance of drawing a heart or a king from a deck.

"At least one six" is everything except "no six"

The complement of an event A, written A′, contains every outcome that is not in A. Since A and A′ together cover the whole sample space without overlapping, P(A) + P(A′) = 1, so P(A′) = 1 − P(A). For one die, P(not a six) = 1 − 1/6 = 5/6. The complement rule is the best tool for "at least one" questions, because "at least one six" is the complement of "no sixes at all", and "no sixes" is easy to compute.

Even or less than four, with the 2 counted once

An outcome is in "A OR B" if it is in A, in B, or in both. Adding P(A) and P(B) counts the outcomes in both twice, so we subtract them once: P(A OR B) = P(A) + P(B) − P(A AND B). OpenStax uses one die with A = even = {2, 4, 6} and B = less than four = {1, 2, 3}. Only 2 is in both. So P(A OR B) = 3/6 + 3/6 − 1/6 = 5/6, which matches counting {1, 2, 3, 4, 6} directly. If two events cannot happen together, they are mutually exclusive, P(A AND B) = 0, and the rule becomes a plain sum.

One die. A = {2, 4, 6} and B = {1, 2, 3} overlap only at 2. The outcome 5 is in neither, so it is the complement of A OR B.

For independent rolls, multiply 5/6 four times

Two events are independent if one happening does not change the chance of the other, like separate rolls of a die or flips of a coin. For independent events, P(A AND B) = P(A) × P(B). Two heads in two flips: 1/2 × 1/2 = 1/4. No six in four rolls: 5/6 × 5/6 × 5/6 × 5/6 = 625/1296. So at least one six in four rolls is 1 − 625/1296 = 671/1296, about 0.518, a little over one half. That is why de Méré won the first bet. When events are not independent, such as drawing cards without putting them back, the multiplication rule needs conditional probability, the subject of the next lesson.

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Four dice at once stand for four throws of one die. The event "At least one 6" is selected, and the readout compares the observed proportion with the exact 671/1296. Roll 10,000 and the two agree to about two decimal places.
Complement ruleP(A′) = 1 − P(A)
Addition ruleP(A OR B) = P(A) + P(B) − P(A AND B)
Mutually exclusiveCannot happen together: P(A AND B) = 0
Independent eventsP(A AND B) = P(A) × P(B)
At least one1 − P(none)
  1. One die. What is P(not a six)?
  2. P(rain) = 0.35. What is P(no rain)?
  3. One die. A = even, B = less than four. What is P(A OR B)?
  4. A standard 52-card deck. What is P(heart OR king)?
  5. Which pair of events is mutually exclusive for one die roll?
    • Even, and greater than 3
    • A 1, and a 6
    • Odd, and less than 4
    • Even, and a 2
  6. Three fair coins. What is P(at least one head)?
  7. At least one six in four rolls of a die, to 3 decimal places?
  8. At least one double six in 24 rolls of two dice, to 3 decimal places?
  9. Independent events with P(A) = 0.3 and P(B) = 0.5. What is P(A AND B)?
  10. Why was de Méré wrong to treat the two bets as equal?
    • Dice were loaded
    • Chances of "at least one" do not grow in proportion to the number of tries
    • Two dice have 30 outcomes
    • The bets had different stakes
Common mistakes
  • "Two dice: P(at least one six) = 1/6 + 1/6 = 1/3." The double six (6, 6) has been counted twice. The addition rule subtracts it once: 1/6 + 1/6 − 1/36 = 11/36, about 0.306. The complement gives the same answer: 1 − (5/6)² = 1 − 25/36 = 11/36.
  • "Six rolls make a six certain: 6 × 1/6 = 1." Multiplying the chance by the number of tries is de Méré's mistake taken further. No six in six rolls has probability (5/6)⁶, about 0.335, so at least one six is about 0.665, not 1.
  • "Mutually exclusive events are independent." It is close to the opposite. On one die, "even" and "odd" cannot happen together, so P(both) = 0, while P(even) × P(odd) = 1/4. Knowing that one happened tells you for certain that the other did not, which is strong dependence.
  • "The complement of 'at least one six' is 'a six every time'." The complement is "no six at all". "A six every time" is one of the results inside "at least one six", not outside it.
Teacher's noteLearned only as formulas, these three rules are easy to mix up. Pair each word with a picture: NOT with shading everything outside a circle, OR with two overlapping circles, AND with the branches of a tree. Once the rules are in place, the de Méré bets make a good test, because the wrong method (scaling 4/6 up to 24/36) feels so natural. In the Korean curriculum these rules belong to the high-school Probability and Statistics course, which also says to avoid complicated problems that assume three or more events are mutually exclusive or independent.
Key points
  • NOT: P(A′) = 1 − P(A). Use it for "at least one".
  • OR: P(A) + P(B) − P(A AND B); mutually exclusive means no overlap.
  • AND for independent events: multiply.
  • One six in 4 rolls ≈ 0.518; double six in 24 rolls ≈ 0.491.
Sources for this course (9)
  • Curriculum: 교육부 고시 제2022-33호 [별책 8] 수학과 교육과정 (ncic.re.kr/inv/org/download.do?year=2022&seq=10003559&orgType=ogi4, PDF read 2026-10-19: 초등학교 5~6학년군 [6수04-04]~[6수04-06] 가능성을 말과 0, 1/2, 1 같은 수로 나타내기; 중학교 1~3학년군 [9수04-03] 상대도수, [9수04-05] 경우의 수(두 경우의 수를 더하거나 곱하는 정도), [9수04-06] 확률(상대도수로서의 의미와 경우의 수의 비율로서의 의미, 각 경우의 가능성이 같다는 가정), 용어 상대도수·사건·확률; 공통수학1 [10공수1-03-01]~[10공수1-03-03] 합의 법칙·곱의 법칙·순열·조합, 기호 nCr; 확률과 통계 [12확통01-03] 이항정리, 용어 이항계수·파스칼의 삼각형; [12확통02-01]~[12확통02-06] 확률의 기본 성질·덧셈정리·여사건의 확률·조건부확률·독립과 종속·곱셈정리, 용어 시행·통계적 확률·수학적 확률·여사건·배반사건·조건부확률·종속·독립·독립시행과 기호 P(A), P(B|A); [12확통03-03] 이항분포, [12확통03-04] 정규분포와 이항분포의 관계, [12확통03-05]·[12확통03-06] 임의추출·표본평균과 모평균, 용어 큰 수의 법칙·이항분포·정규분포·모평균·표본평균과 기호 B(n, p), N(m, σ²), X̄)
  • Textbook: OpenStax, Introductory Statistics 2e by Barbara Illowsky and Susan Dean, 3.1 terminology, equally likely outcomes, dime and nickel, law of large numbers, Belgian one euro coin 56% heads in 250 trials, conditional probability, Table 3.1 handedness of 100 people (openstax.org/books/introductory-statistics-2e/pages/3-1-terminology)
  • 4.3 binomial distribution
  • 7.1 central limit theorem for sample means, example mean 90, SD 15, n = 25, standard error 3 (.../pages/7-1-the-central-limit-theorem-for-sample-means-averages)
  • Yut rules for the Korean example only: 국립민속박물관 한국민속대백과사전 「윷놀이」, 도·개·걸·윷·모 by how many of the four sticks land 뒤집어진 (folkency.nfm.go.kr/kr/topic/detail/1595, read 2026-10-19)
  • Kerrich story: John Kerrich, 10,000 tosses, 5,067 heads (JMP case study JMP014 by DeWayne Derryberry, jmp.com/en/academic/case-study-library/is-a-coin-fair, citing Baldi and Moore; Wikipedia "John Edmund Kerrich")
  • Galileo story: Sopra le scoperte dei dadi, sum 10 in 27 ways and 9 in 25 ways with three dice (English translation, University of York history of statistics, york.ac.uk/depts/maths/histstat/galileo.pdf)
  • Pascal story: Pascal to Fermat, 29 July 1654, one six in four throws "as 671 to 625" (American Physical Society, aps.org/apsnews/2009/07/pascal-letters-fermat-points; Gaston Sanchez, De Mere's games; Statistical Science 2024 "Antoine Gombaud, Chevalier de Méré")
  • Galton story: Francis Galton, Natural Inheritance of 1889, the quincunx (Wikipedia "Galton board")

Checked: 2026-10-17